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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

PhysicsMediumsingle choice

A particle has an initial velocity of $12 m / s$ due east and a constant acceleration of $4 m / s^{2}$ due west. The distance covered by the particle in the fourth second of its motion is:

Options:

Answer:

Solution:

It comes to rest at $t = \frac{12}{4} = 3$ s. During the whole fourth second it moves back from rest with acceleration $4 m / s^{2}$ , covering $\frac{1}{2} (4) 1^{2} = 2$ m. Option B.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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